how to find the next minimum value using if..else only?

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aditya sahu
aditya sahu on 16 May 2017
Commented: aditya sahu on 17 May 2017
let the given values are, b1(k)=195, b1(k+1)=190,and p1(k)=204,p1(k+1)=187,
I need to find the optimal value of p1(k) and p1(k+1).
ax= abs(b1(k)-p1(k)) + abs(b1(k+1)-p11(k+1))= 22
bx= abs(b1(k)-p1(k)) + abs(b1(k+1)-p111(k+1))=28
cx= abs(b1(k)-p11(k)) + abs(b1(k+1)-p1(k+1))=28
dx= abs(b1(k)-p11(k)) + abs(b1(k+1)-p11(k+1))=55
ex= abs(b1(k)-p11(k)) + abs(b1(k+1)-p111(k+1))=44
fx=abs(b1(k)-p111(k)) + abs(b1(k+1)-p1(k+1))=10
gx=abs(b1(k)-p111(k)) + abs(b1(k+1)-p11(k+1))=20
hx=abs(b1(k)-p111(k)) + abs(b1(k+1)-p111(k+1))=26
I need to find the optimal value of p1(k) and p1(k+1) out of available 8 combinations with 3 conditions. The 8 combinations are
1. p1(k),p11(k+1)
2. p1(k),p111(K+1)
3. p11(k),p1(k+1)
4. p11(k),p11(k+1)
6. p11(k),p111(k+1)
7. p111(k),p1(k+1)
8. p111(k),p11(k+1)
9. p111(k),p111(k+1).
and the 3 conditions are
1. The absolute difference between p1(k),p1(k+1) must be between 0 to 15.
2. Any one value out of p1(k) or p1(k+1) must be greater than 192.
3. I should consider first the minimum value among 'ax' to 'hx'
The actual probelem i am facing is, when i choose the smallest value 'fx'=10 then its corresponding values are p1(k)=p111(k)=188 and
p1(k+1)=p1(k+1)=187 ,,,here both the values are less than 192.i dont know how to check for the next lowest value i.e 'gx'=20.
aditya sahu
aditya sahu on 17 May 2017
Dear sir, it is also ok by using any techhnique.

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